Symmetric matrices and dot products

Transcription

1 Symmetric matrices and dot products Proposition An n n matrix A is symmetric iff, for all x, y in R n, (Ax) y = x (Ay). Proof. If A is symmetric, then (Ax) y = x T A T y = x T Ay = x (Ay). If equality holds for all x, y in R n, let x, y vary over the standard basis of R n. Corollary If A is symmetric and x, y are eigvecs corresponding to different eigvals λ, µ, then x y = 0. Proof. λx y = (Ax) y = x (Ay) = x (µy) = µx y, so (λ µ)(x y) = 0, but λ µ 0.

2 Proposition Every eigenvalue of a symmetric matrix (with real entries) is real, and we can pick the corresponding eigenvector to have real entries. Proof. Suppose A is symmetric and Ax = λx with x 0 (maybe with complex entries). Then Ax = Ax = λx, so λ(x x) = (Ax) x = x (Ax) = λ(x x) ; but x x is real and positive, so λ = λ. And if x is an eigvec with complex entries corr to λ, then we can multiply by some complex number and get at least one real nonzero entry, and then x + x is an eigvec corr to λ with real entries.

3 Lemma If A is a symmetric matrix with (real) eigvec x, then if y is perpendicular to x, then so is Ay (i.e., A multiplies the orthogonal complement (Rx) into itself). Proof. Let λ be the corr eigval. Because y x = 0, we have (Ay) x = y (Ax) = y (λx) = λ(y x) = 0.

4 Symmetrics have orthogonal diagonalization Theorem (Spectral Theorem) If A is symmetric, then there is an orthogonal matrix Q and a diagonal matrix Λ for which A = QΛQ T. Proof Let λ 1, x 1 be a real eigval and eigvec of A, with x = 1. Pick an orthonormal basis B for (Rx) ; then x, B form an orthonormal basis for R n. Use them as columns of an orthogonal matrix Q 1, with x 1 first; then because A multiplies (Rx) into itself, we have A = Q 1 [ λ1 0 0 A 2 Q T 1.

5 Because Q 1 [ λ1 0 0 A 2 Q T 1 = A = A T = Q 1 [ λ1 0 0 A T 2 Q T 1, A 2 is still symmetric; so we can repeat the process with A 2 ; and if [ λ2 0 A 2 = Q 2 Q2 T 0 A 3 then A = Q 1 [ Q 2 λ λ A 3 [ Q2 T Q T 1. Continuing, we get to the desired form.

6 Example For a symmetic A, R gives orthonormal eigenvectors (and eigenvalues in decreasing order).

7 Suppose we have A symmetric, Q = [ x 1... x n and Λ = diag(λ 1,..., λ n ) with A = QΛQ T. Write λ Λ = λ n

8 Then λ Q QT λ = [ x 1 x 2... x n x T 1 x T 2. x T n = λ 1 x 1 x T 1

9 And so on, so we get A = λ 1 x 1 x T 1 + λ 2 x 2 x T λ n x n x T n. The x i x T i s are the projection matrices P i s on the subspaces Rx i s. (And they are rank 1 matrices.) Because they are the orthogonal projections onto an orthonormal basis, they add up to I if we apply them all to the same vector and add up the results, we ll get back the original vector.

10 Example: [ 9 2 A = 2 6 : λ 1 = 5, x 1 = [ 1/ 5 2/ 5 λ 2 = 10, x 2 = [ 2/ 5 1/ 5 and [ 1/5 2/5 2/5 4/5 [ 1/ 5 [ A = 5 2/ 1/ 5 2/ 5 5 [ 2/ 5 [ / 5 1/ 5 [ 1/5 2/5 = 5 2/5 4/5 [ + 1/ 5 [ /5 2/5 2/5 1/5 4/5 2/5 2/5 1/5 = I.

11 Definition A quadratic form on R n is a function q : R n R given by a polynomial in which every term has degree 2: q(x) = i,j a i,jx i x j. Because x i x j = x j x i, we can assume a i,j = a j,i, and rewrite q(x) = x T Ax where A is symmetric. If q(x) > 0 for every nonzero x, then q and A are positive definite. If q(x) 0 for every x, then they are positive semidefinite.

12 Example 1 x x2 2 3x3 2 4x 1 x 2 + x 1 x 3 + 2x 2 x 3 = x T 1 2 1/ /2 1 3 Because of the 3 on the main diagonal, this quadratic form is not even pos semidef (nor is its associated matrix): [ / / = 3 Moral: If any main diagonal entry is 0, the matrix is not pos def. x

13 Example 2 Positive definite: (x + 3y) 2 + 4(2x y) 2 = 17x 2 10xy + 13y 2 = [ x y [ [ x y Example 3 (x + 3y) 2 + 4(2y z) 2 + 2(2x + 3z) 2 = 9x y z 2 + 6xy 16yz + 24xz = [ x y z x y z Certainly always nonnegative, but if x = 3y = 3z/2, all 3 squares are 0; so pos semidef.

14 Example 4 What about f (x, y, z) = 23x y 2 + 8z 2 28xy + 4xz + 32yz? f (x, y) = 23(x xz z2 ) + 14(y yz z2 ) ( ) z 2 28xy 7 = 23(x z)2 + 14(y 8 7 z) z2 28xy. Can we make this negative for some x, y, z? Well, try setting x = (2/23)z and y = (8/7)z: That will make f into (1652/161)z 2, which is negative for any z 0. So f is not even pos semidef.

15 Application: Second Derivative Test Suppose z = f (x, y) is a function of two variables, and at a point (a, b) the first derivatives f / x and f / y are both 0. Then the Taylor series for f at that point is f (a, b) + 2 f x 2 (a, b)(x a)2 + 2 f (a, b)(x a)(y b) x y + 2 f y x (a, b)(y b)(x a) + 2 f y 2 (a, b)(y b) where the rest is higher-degree terms. So near (a, b), f behaves like the quadratic part, which is a quadratic form. When this form is positive definite, the smallest value f takes on in a small neighborhood of (a, b) at (a, b). In other words, (a, b) is a local minimum of f.

16 So it would be useful to have a quick way to decide whether the matrix is positive definite: [ 2 f (a, b) 2 f (a, b) 2 x 2 f y x (a, b) x y 2 f y 2 (a, b)

17 Theorem Let A be a symmetric matrix. TFAE: (a) All pivots of A are positive. (b) All upper left subdeterminants are positive. (This is Sylvester s criterion.) (c) All eigenvalues of A are positive. (d) The quadratic form x T Ax is positive definite. (This is the energy-based definition.) (e) A = R T R where R has independent columns. Proof (a) = (b): a 1,1 is first pivot, hence pos, and first subdet. By row ops, upper left subdets of A are equal to subdets of a matrix, say B, with the rest of the first column 0 s. Second entry on B s main diagonal is A s 2nd pivot; because it and a 1,1 are pos, B s 2nd upper left subdet is also pos. Etc.

18 (b) = (a): First subdet is a 1,1 > 0, first pivot. Repeat the rest of (a) = (b), except reversing causality: k-th pivot is pos because k k subdet is pos. (c) (d): Write A = QΛQ T ; then x T Ax = x T QΛQ T x = (Q T x) T Λ(Q T x) = i λ i y 2 i, where the y i s are the entries of Q T x. So if all the λ i s are positive, x T Ax is positive definite; while if one is negative, then there are x s for which x T Ax is negative. (c) = (e): Write A = QΛQ T. Then by (c), Λ makes sense and is invertible. Set R = Q ΛQ T. Then A = R T R, and the columns of R are independent. (e) = (d): Given A = R T R, x T Ax = (Rx) (Rx) = Rx 2 0. And if R has ind cols, then the only x with Rx 2 = 0 is x = 0.

19 (a) = (e): Because A is symmetric, (a) says it has an LDU-decomposition A = LDL T where L is lower triangular with 1 s on the main diagonal, and D is diagonal with the pivots of A on its diagonal. Set R = DL T. (d) = (b): Let b k = the upper left k k subdet of A, and c kj = the cofactor of a kj if we evaluate b k along its last row. Then for i < k, k j=1 a ijc kj is the value of the same det as b k except that its last row is equal to its i-th row, so it is 0; but k j=1 a kjc kj = b k. Note c kk = b k 1. We prove by induction (and pos defness) that all b k s are pos: Because A is pos def, b 1 = a 11 > 0. Assume b k 1 > 0. Set Then so b k > 0. x = (c k1, c k2,..., c kk, 0,..., 0). 0 < x T Ax = x T (0,..., 0, b k,?,...,?) = b k 1 b k,

20 Notes on proof: (e): The proof of (c) = (e) gives one square, in fact symmetric, R that works (the Cholesky decomposition), but any R with independent columns, even if not square, works. Example: A = R T R = [ = [ has independent columns, so A is positive definite. Therefore, the eigenvalues of A are all negative, so the solutions of du dt = Au approach 0 as t., and R (c): The (c) = (d) part of the proof shows us the right way to complete squares in quadratic forms. (Example, next page.)

21 Example 4, revisited: For f (x, y, z) = 23x y 2 + 8z 2 28xy + 4xz + 32yz, the matrix is A = = QΛQ T where Q = 1/3 2/3 2/3 2/3 1/3 2/3 2/3 2/3 1/3 and Λ = , so x T Ax = x T QΛQ T x = (Q T x) T Λ(Q T x) = 9( 1 3 x y 2 3 z)2 + 18( 2 3 x y z)2 + 36( 2 3 x y z)2 ; again, not even pos semidef.

22 Proposition A symmetric matrix is positive semidefinite iff its eigenvalues are nonnegative, or equivalently iff it has the form R T R. Proof. Same as the positive definite case, except that, if the cols of R are not ind, there are nonzero x s for which Rx = 0.

23 Example 1 revisited: [ det[1 = 1, det x T / /2 1 3 = 2, det x 1 2 1/ /2 1 3 = 5/2 Not even positive semidefinite. Eigenvalues from R: , ,

24 Example 2 revisited: Positive definite: [ x y [ [ x y [ 17 5 det[17 = 17, det 5 13 Eigenvalues from R: , = 196

25 Example 3 revisited: det[9 = 9, (x + 3y) 2 + 4(2y z) 2 + 2(2x + 3z) 2 = 9x y z 2 + 6xy 16yz + 24xz = [ x y z x y z [ 9 3 det 3 25 = 216, det pos semidef. (Eigvals from R: , , 0.) = 0 ;

26 Example 4 re-revisited: 23x y 2 + 8z 2 28xy + 4xz + 32yz = x T Ax where A = det[23 = 23, [ det = 126, det A = 5832, so not pos def. (We saw eigvals were 9, 18, 36.)

27 Application: Tilted Ellipses The equation of an ellipse in standard position is x 2 /a 2 + y 2 /b 2 = 1, i.e., [ x y [ 1/a /b 2 [ x y = 1. The semiaxes are a and b (the positive square roots of the reciprocals of the eigvals). But a rotation matrix Q gives a tilted ellipse. The axes are in the directions of the columns of Q.

28 Example 5 [ 1/16 0 Λ = 0 1/9 A = QΛQ T = [ 4/5 3/5, Q = 3/5 4/5 [ 288/ / / /3600 : 1 16 x y 2 = x xy y 2 = 1

29 Example 6 Describe the ellipse 13x 2 6 3xy + 7y 2 = 4. Divide by 4: 13 4 x xy y 2 = 1 The quadratic form has matrix [ 13/4 3 3/4 A = 3 = QΛQ T 3/4 7/4 where [ Q = 1/2 3/2 3/2 1/2, Λ = [ So the ellipse has axes tilted 60 from the xy-axes, with half-lengths 1 and 1/2..

30 Example 7 Describe the ellipse x 2 + 6xy + 4y 2 = 16. The matrix of the quadratic form is [ 1/16 3/16 A = 3/16 1/4, and R shows that its eigvals are and The negative eigenvalue means that it is not an ellipse. The graph of the quadratic form z = (x 2 + 6xy + 4y 2 )/16 is a saddle, and the cross section z = 1 is a hyperbola. Or, [ 1/16 3/16 det[1/16 = 1/16, det 3/16 1/4 = 5/256 < 0